Advertisements
Advertisements
प्रश्न
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
Advertisements
उत्तर
Given: Lowering in melting point (ΔTf) = 1.5°C
Mass of solvent (CH3COOH), w1 = 75 g
Mass of solute, w2 = ?
Molar mass of solvent (CH3COOH), M1 = 60 g mol−1
Molar mass of solute (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16
= 72 + 8 + 96
= 176 g mol−1
Kf = 3.9 K kg mol−1
Applying the formula,
ΔTf = `(K_f xx w_2 xx 1000)/(M_2 xx w_1)`
⇒ w2 = `(Delta T_f xx M_2 xx w_1)/(K_f xx 1000)`
= `(1.5 xx 176 xx 75)/(3.9 xx 1000)`
= 5.08 g (approx)
Hence, 5.08 g of ascorbic acid needs to be dissolved.
APPEARS IN
संबंधित प्रश्न
1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. 1.6x10 Kg of another non-electrolyte solute when dissolved in 0.086 Kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol−1. Calculate the atomic masses of A and B.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–1)
Pure benzene freezes at 5.45°C. A 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene is:
0.01 M solution of KCl and BaCl2 are prepared in water. The freezing point of KCl is found to be – 2°C. What is the freezing point of BaCl2 to be completely ionised?
Which of the following 0.10 m aqueous solutions will have the lowest freezing point?
A solution containing 1.8 g of a compound (empirical formula CH2O) in 40 g of water is observed to freeze at –0.465° C. The molecular formula of the compound is (Kf of water = 1.86 kg K mol–1):
If molality of dilute solution is doubled, the value of molal depression constant (Kf) will be ______.
Given below are two statements labelled as Assertion (A) and Reason (R).
Assertion (A): Cryoscopic constant depends on nature of solvent.
Reason (R): Cryoscopic constant is a universal constant.
Select the most appropriate answer from the options given below:
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
Assertion: When NaCl is added to water a depression in freezing point is observed.
Reason: The lowering of vapour pressure of a solution causes depression in the freezing point.
Calculate freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at – 0.320°C, what will be the value of van't Hoff factor? (kg for water = 108b°C mol–1)
When 25.6 g of sulphur was dissolved in 1000 g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of sulphur (Sr).
(Kf for benzene = 5.12 K kg mol−1, Atomic mass of sulphur = 32 g mol−1)
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
The of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
